## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 388: 117

#### Answer

$$\frac{1}{6}$$

#### Work Step by Step

Since \begin{align*} \lim _{y \rightarrow 0} \frac{\sin ^{-1} y-y}{y^{3}} &=\frac{0}{0} \end{align*} Then by using L’Hôpital’s Rule \begin{align*} \lim _{y \rightarrow 0} \frac{\sin ^{-1} y-y}{y^{3}}&=\lim _{y \rightarrow 0} \frac{\frac{1}{\sqrt{1-y^2}} -1}{3y^{2}}\\ &=\lim _{y\to \:0}\left(\frac{1-\sqrt{1-y^2}}{3y^2\sqrt{1-y^2}}\right)\\ &= \lim _{y\to \:0}\left(\frac{\frac{y}{\sqrt{-y^2+1}}}{\frac{3\left(-3y^3+2y\right)}{\sqrt{1-y^2}}}\right)\\ &= \lim _{y\to \:0}\left(\frac{1}{3\left(-3y^2+2\right)}\right)\\ &=\frac{1}{6} \end{align*}

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